Issue
Advise, how to press search button using espresso on the keyboard for after typing text request
//Fragment class
@Override
public void onCreateOptionsMenu(Menu menu, MenuInflater inflater) {
inflater.inflate(R.menu.search_menu, menu);
searchMenuItem = menu.findItem(R.id.action_search);
if (searchMenuItem != null) {
searchView = (SearchView) MenuItemCompat.getActionView(searchMenuItem);
searchView.setOnQueryTextListener(this);
searchView.setQueryHint("Search");
}
super.onCreateOptionsMenu(menu, inflater);
}
class from androidTest is:
public class GridFragmentTest {
private final static String TAG = GridFragmentTest.class.getSimpleName();
@Rule
public ActivityTestRule activityTestRule = new ActivityTestRule(MainActivity.class);
@Test
public void GridFragmentRecycleViewTest(){
IdlingResource componentIdlingResource = getIdlingResource();
Espresso.registerIdlingResources(componentIdlingResource);
Log.d(TAG, "GridFragmentRecycleViewTest()");
onView(withId(R.id.action_search)).perform(click());
onView(isAssignableFrom(SearchView.class)).perform(typeSearchViewText("lord")).perform(pressImeActionButton());
EspressoIdlingResource.increment();
onView(withId(R.id.recycler_view)).check(new RecyclerViewItemCountAssertion(10));
}
public static ViewAction typeSearchViewText(final String text){
return new ViewAction(){
@Override
public Matcher<View> getConstraints() {
//Ensure that only apply if it is a SearchView and if it is visible.
return allOf(isDisplayed(), isAssignableFrom(SearchView.class));
}
@Override
public String getDescription() {
return "Change view text";
}
@Override
public void perform(UiController uiController, View view) {
((SearchView) view).setQuery(text,false);
}
};
}
}
The is a PerformException for the code:
.perform(pressImeActionButton()
Exception text is:
androidx.test.espresso.PerformException: Error performing 'androidx.test.espresso.action.EditorAction@9c9ab29' on view 'is assignable from class: class androidx.appcompat.widget.SearchView'.
at androidx.test.espresso.PerformException$Builder.build(PerformException.java:84)
at androidx.test.espresso.base.DefaultFailureHandler.getUserFriendlyError(DefaultFailureHandler.java:85)
at androidx.test.espresso.base.DefaultFailureHandler.handle(DefaultFailureHandler.java:57)
at androidx.test.espresso.ViewInteraction.waitForAndHandleInteractionResults(ViewInteraction.java:316)
at androidx.test.espresso.ViewInteraction.desugaredPerform(ViewInteraction.java:177)
at androidx.test.espresso.ViewInteraction.perform(ViewInteraction.java:118)
Solution
to get the view, you can use the resources.getSystem().getIdentifier method to get the search text field, after that apply the following methods
clearText() typeText("bla bla bla")
onView(withId(Resources.getSystem().getIdentifier("search_src_text",
"id", "android"))).perform(clearText(),typeText("enter the text"))
.perform(pressKey(KeyEvent.KEYCODE_ENTER))
Answered By - Ahmed Garhy
0 comments:
Post a Comment
Note: Only a member of this blog may post a comment.